The two predominant topologies used for class D amplifiers are the full-bridge- and the half-bridge topology. A prior art setup of these two topologies including power supply is shown in FIG. 2 and FIG. 3.
For low cost consumer products the half-bridge topology has a distinct advantage over the full-bridge since only half the number of power switches, drivers and output inductors is needed. Even though the voltage rating of the power switches is doubled in the half-bridge topology, there is still a manufacturing and cost benefit of the reduction. Since the speaker output terminals are referenced to ground, overload protection and output sensing becomes easier compared to the full-bridge topology.
The drawback of the half-bridge class D topology is the need for a dual rail supply and the fact that this topology will pump current back to the opposite supply rail from where the power is being drained. As shown in FIG. 4 the negative rail capacitor C2 will be charged by the output inductor current during the interval tε[DT;T].
The charge delivered to back to the rail is dependent on load impedance, audio frequency, power level and rail voltage. These dependencies can be expressed as:
                              Δ          ⁢                                          ⁢          Q                =                                            ∫              0              Audio                        ⁢                                                            V                  Peak                                ·                                  sin                  ⁡                                      (                                          2                      ·                      π                      ·                                              f                        Audio                                            ·                                        )                                                                              2                ·                                  R                  Load                                                              -                                                                      (                                                            V                      Peak                                        ·                                          sin                      ⁡                                              (                                                  2                          ·                          π                          ·                                                      f                            Audio                                                    ·                          t                                                )                                                                              )                                2                                            2                ·                                  R                  Load                                ·                                  V                  Rail                                                      ·                          ⅆ              t                                                          (        1        )                                                          ⁢                              Δ            ⁢                                                  ⁢            Q                    =                                                    2                ·                                  V                  Audio                                            -                                                π                  ·                                      V                    Audio                    2                                                                    2                  ·                                      V                    Rail                                                                                      4              ·              π              ·                              f                Audio                            ·                              R                Load                                                                        (        2        )            
The worst case pumping can be found by solving dQ/dV Audio=0 for (2):
                                          ⅆ            Q                                ⅆ                          V              Audio                                      =                                                            2                ·                                  V                  Rail                                            -                              π                ·                                  V                  Audio                                                                    2              ·              π              ·              f              ·                              V                Rail                                              =                                    0              ⇔                              V                Audio                                      =                                          2                π                            ·                              V                Rail                                                                        (        3        )            
The worst case charge can then be found by inserting (3) into (2):
                              Δ          ⁢                                          ⁢                      Q            Max                          =                              V            Rail                                2            ·                          π              2                        ·                          f              Audio                        ·                          R              Load                                                          (        4        )            
Capacitor minimum size for a given rail voltage rise (a fraction of nominal rail voltage) can then be found as:
                              C          Min                =                  1                                    2              ·                              π                2                            ·                              f                Audio                            ·              Δ                        ⁢                                                  ⁢                          V              relative                                                          (        5        )            
FIG. 5 shows an example with a 20 Hz audio signal and an 4 ohm load impedance where the minimum rail capacitance is determined as a function of the allowable rail perturbation. If we allow for a 20% increase in rail voltage the minimum capacitance/channel/rail is approximately 3000 uF. For a stereo setup with a single-ended class D amplifier, the requirement win be 4 pcs 3300 uF capacitors to keep rail voltages within 20%. The resulting rail variation for the above example is shown in FIG. 5. The nominal rail voltage is 40 VDC and the perturbed rail is increased with approximately 8V (20%). The audio signal is equal to 2/π of the nominal rail voltage which is the worst case situation.
Also performance of the amplifier will be affected by the rail pumping especially non-feedback class D amplifiers where the asymmetrical rail perturbation will be modulated directly onto the audio signal dramatically increasing the distortion. Because of the dual supply with asymmetrical perturbation, any feed-forward system implemented will have little or no effect.
The benefits in terms of component count and complexity of using the single-ended class D amplifier compared to the full-bridge class D amplifier are quickly disappearing when taking into account the problems introduced by the rail pumping phenomena discussed above. These drawbacks can be summarized as:                Excessive use of rail capacitors=>higher cost and larger physical size        Higher voltage-rated devices in terms of rail Capacitors and amplifier MOSFETs=>higher cost and larger physical size        Unacceptable audio performance for non-feedback class D amplifiers and degraded audio performance for feedback class D amplifiers=>reduced product value        
A voltage equalization scheme for a single-ended class D amplifier is suggested in WO 2005/091497. This circuit suffers from several disadvantages not encountered in the invention. Some of the disadvantages are:
1) The circuit redistributes the pumping energy in a 2 step operation. In the first step energy is stored in an inductor to be released in the second step.
2) The circuit uses two separate control systems.
3) The circuit will only limit the rail pumping—not cancel it.
4) The circuit will not help utilizing rail capacitance on both rails.
5) The circuit is based on a “buck-boost” topology renowned for its poor conversion efficiency.
If the rail pumping is to be effectively cancelled, a not insignificant amount of power has to be recycled between the rails. From (4) we can calculate the worst case average power that has to be converted from one rail to the other:
                              P          Av_Pumping                =                              V            Rail            2                                              π              2                        ·                          R              Load                                                          (        6        )            
Introducing the maximum modulation index, MMax, as the fraction of maximum output voltage to rail voltage, the relation between average pumping power and maximum output power can be found as:
                                          P            Av_Pumping                                P            Max_Rload                          =                  2                                    π              2                        ·                          M              Max              2                                                          (        7        )            
In a stereo setup of 2×125 W in 4 ohm with rail voltages of +/−40V and a maximum modulation index of 0.85, the average power to be redistributed in the worst case situation is approximately 81 W which accounts for 32% of the power delivered to the load!
To benefit from the advantages of using the single-ended class D amplifier the rail pumping problem needs to be addressed in an intelligent manner adding minimum complexity and cost to the audio power conversion system.